Question

$\begin{array}{cccc}& \text{Column I}& & \text{Column II}\\ \text{(a)}& arg\left(\frac{z+1}{z-1}\right)=\frac{\pi }{4}& \text{(p)}& \text{Parabola}\\ \text{(b)}& z=\frac{3i-1}{2+it}\left(tϵR\right)& \text{(q)}& \text{Part of a circle}\\ \text{(c)}& argz=\frac{\pi }{4}& \text{(r)}& \text{Full circle}\\ \text{(d)}& z=t+i{t}^2\left(tϵR\right)& \text{(s)}& \text{Line}\end{array}$

Which of the following is correct?

• A. A-Q,B-R,C-S,D-Q
• B. A-S,B-R,C-P,D-Q
• C. A-Q,B-R,C-S,D-P
• D. A-R,B-S,C-P,D-Q

Answer 1

The correct option is C

A-Q,B-R,C-S,D-P

$arg\left(\frac{z+1}{z-1}\right)=\frac{\pi }{4}$

$Letz=x+iy$

$\Rightarrow arg\left(\frac{x+iy+1}{x+iy-1}\right)=\frac{\pi }{4}$

$\Rightarrow arg\left(\frac{(x+1)+iy}{(x-1)+iy}\right)=\frac{\pi }{4}$

$\Rightarrow arg\left(\frac{(x^2+y^2-1)+i(-2y)}{(x-1{)}^2+y^2}\right)=\frac{\pi }{4}$

$\Rightarrow \left(\frac{-2y}{x^2+y^2-1}\right)=\frac{\pi }{4};-2y>0\&x^2+y^2-1>0$

$\Rightarrow \frac{-2y}{x^2+y^2-1}=1$

$\Rightarrow x^2(y+1{)}^2=2;\text{which is a fall circle}$

$but\because -2y>0\Rightarrow y<0;$

hence this is only part of a circle.

$\left(b\right)z=\frac{3i-1}{z+it}\left(tϵR\right)$

Let z = x + iy

$x+iy=\frac{-1+3i}{z+it}$

$=\frac{(3t-2)}{t^2+4}+i\frac{t+6}{t^2+4}$

$\Rightarrow x=\frac{3r-2}{t^2+4}$

$;y=\frac{t+6}{t^2+4}$

$\Rightarrow x^2=\frac{9t^2+4-12t}{(t^2+4{)}^2};y^2=\frac{36+t^2+12t}{(t^2+4{)}^2}$

$\therefore x^2+y^2=\frac{{10}^2+40}{(t^2+4^2)}=\frac{10}{t^2+4}......\left(1\right)$

$x-3y=\frac{-2-18}{t^2+4}=\frac{-20}{t^2+4}=\frac{-20}{t^{+}4}......\left(2\right)$

$\frac{x-3y}{-20}=\frac{x^2+y^2}{10}$

$x^2+y^2+\frac{x}{2}-\frac{3y}{2}=0⟶\text{Full circle}$

$\left(c\right)argz=\frac{\pi }{4}$

z is a line passing through origin at $\frac{\pi }{4}$ angle with X - axis

$\left(d\right)z=t+i{t}^2\left(tϵR\right)$

$x+iy=t+i{t}^2$

$x=t;t+i{t}^2$

$\Rightarrow y=x^2⟶\text{Parabola}$